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Offline Farsight

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« Reply #50 on: 15/02/2010 14:27:47 »
Interesting article graham, thanks. There's some interesting comments re Einstein and gravity, about point particles, and even the near field aka evanescent wave:

Much of the literature on the question of radiation from accelerating charges focuses on the Lorentz-Dirac equation of motion for a classical charged point-like particle interacting both with an external field and with its own field. This equation is the source of equation (1), but it's important to remember that it is based on classical electrodynamics of point-like particles, rather than on quantum electrodynamics, so it's physical relevance is questionable. Moreover, the internal validity of the Lorentz-Dirac equation is cast into doubt by the existence of run-away solutions (see below)...

In addition, the usual neglect of the "near" static field when dealing with the "far" radiation field...


 

Offline Ron Hughes

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« Reply #51 on: 15/02/2010 14:58:34 »
If the velocity and acceleration referred to in Einstein's Relativity is correct then the paradox is resolved. I will post what I consider to be an explanation of the double slit,both the photon and the electron, sometime this afternoon or tonight here    http://www.thenakedscientists.com/forum/index.php?topic=28667.msg299038#msg299038
« Last Edit: 15/02/2010 15:02:58 by Ron Hughes »
 

Offline graham.d

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Offline Ron Hughes

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« Reply #53 on: 16/02/2010 17:20:25 »
Good link Graham. His paper agrees with my own view.
 

Offline graham.d

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« Reply #54 on: 17/02/2010 14:28:54 »
The simple answer to my question, which now seems obvious, is that a charged particle in a circular orbit does not emit radiation. Even looking at this classically, you can consider a continuous line of charges in orbit. This would be the same as an electric current and would produce a constant magnetic field, but not em radiation. The only energy used is in establishing the field i.e. getting the electrons into orbit.

The first proposed question about being in free fall, but not in orbit. i.e. following a hyperbolic path or simply falling to the ground, is more complicated. By hyperbolic here I mean simply a hyperbolic path in Euclidean space and not Hyperbolic acceleration (Contant Proper Acceleration) as in the paper a cited. I don't think the paper fully answers this directly, though it needs much study to grasp all the concepts and to appreciate the comments on the large number of differing views on this that are presented. It seems that a charged particle in free fall will fall equally fast as a non-charged particle, but will radiate (according to Rohrlich). The energy he attributes being supplied by the "Schott Acceleration Energy". I have not come across this before and would like to get a better understanding when time allows. The maths and concepts are not trivial. The paper suggests that Schott energy is located locally around the particle but does not contribute to its rest mass (which would be really troubling).

I am unclear whether a comoving observer would see the radiation. I don't think so. But this would not necessarily be ruled out on the basis of equivalence because there has to be boundary conditions which may result in there having to be a converging gravity field. I'm not sure about this.
 

Offline Ron Hughes

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« Reply #55 on: 17/02/2010 16:53:02 »
" The paper suggests that Schott energy is located locally around the particle but does not contribute to its rest mass (which would be really troubling). " Yes Graham, I would find it extremely troubling. Later I will dig into that a little more.
 

Offline yor_on

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« Reply #56 on: 17/02/2010 20:42:09 »
An electron in orbit around a gravitating body is accelerating. In Classical terms, inwards toward the centre of the body at a rate of (v^2)/r. From my distant position I see that electron accelerating due to the force of gravity exerted by the body. If it is accelerating will it emit radiation? If so will it's orbit decay? If it's orbit decays then if you are orbiting with the electron I will see the electron orbit decay but you stay in your orbit. Now, the fact that we are different observers cannot account for two different events. From your co-orbit perspective why would the electron move away from you as you are both in free fall and following the same geodesic?

This "An electron in orbit around a gravitating body is accelerating. In Classical terms, inwards toward the centre of the body at a rate of (v^2)/r."

Would that be a uniformly accelerating velocity Graham?
Both this and mine free falling particle seems to fall outside that scenario if we by 'uniformly accelerating' means accelerating at a constant 'gravity' within its own frame of reference, as f.ex at one G.?

And if our cases won't meet that 'prerequisite' I think what we are discussing is a different scenario from Ron's?
Or am I wrong here??

Why I think so is because what Ron's scenario describes as well as Antiphons seems to be a universe from the 'Rindler observers' perspective 'locally'?
==

Okay I saw that you already answered that one. Sorry, missed it Graham..
« Last Edit: 19/02/2010 07:52:20 by yor_on »
 

Offline Ron Hughes

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« Reply #57 on: 17/02/2010 20:54:24 »
The point is this. An observer that is stationary in space far from the gravitating body will see the electron emit radiation. An observer that is at the center of the gravitating body will not see the electron emit radiation. This scenario fits perfectly with GR.
 

Offline graham.d

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« Reply #58 on: 18/02/2010 09:02:58 »
A charged body in orbit will not radiate, but a charged body that is in some non-periodic motion (falling or on a hyperbolic path) will, I think, according to Eriksen's paper. The Larmor formula seems a special case and not universally applicable.

Ron, I don't see how you reach that conclusion. Nobody has been speaking about an observer at the centre of the gravitating body. And it is not 100% clear, given the eminent physicists in disagreement, that the theory does all fit with GR. The maths, at present, results in an asymmetry with time and a non-causal pre-acceleration of a body in certain cases of abrupt directional changes. These results are cause for some concern at least.
 

Offline Ron Hughes

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« Reply #59 on: 18/02/2010 14:06:44 »
I think but I could be wrong that for every scientist you find that does not believe GR I can find ten thousand that do.
 

Offline graham.d

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« Reply #60 on: 18/02/2010 16:55:30 »
It is not that I don't "believe" in GR, Ron. Generally I do. But I do not understand the action of an accelerating charge in a gravity field. And, it seems surprisingly, neither does anyone else, although they clearly understand it a lot more than me. Eriksen's paper is not definitive but is valuable because he quotes the conflicting results that have been presented by lots of people including Pauli(!), Unruh, Rohrlich, Teitelboim etc. plus commentary on their work by others. I have read the paper a few times but not enough to grasp all of it and I find the maths very hard with some Tensor notations that I am not familiar with; also I have never used tensors much for about 35 years so to say I'm rusty would be a severe understatement. Not to mention the fact that the paper quotes equations without defining the terms, which does not help a lot. I am trying though :-) The results are really quite counter-intuitive.
 

Offline Ron Hughes

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« Reply #61 on: 18/02/2010 17:13:55 »
Maybe I don't fully understand the question. An electron falling toward a star will emit radiation to all observers except an observer falling with the electron. An electron orbiting a star will emit radiation to a far away observer stationary with respect to the star but to an observer at the center of the star.
 

Offline graham.d

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« Reply #62 on: 18/02/2010 17:46:20 »
Maybe I don't fully understand the question. An electron falling toward a star will emit radiation to all observers except an observer falling with the electron. An electron orbiting a star will emit radiation to a far away observer stationary with respect to the star but to an observer at the center of the star.

I guess you intended "[...] with respect to the star but NOT to an observer at the center of the star." [...]

otherwise the words don't make sense.

An electron falling toward a star will emit radiation... True according to recent papers but not what Pauli thought in 1911.

An electron falling toward a star will emit radiation to all observers except an observer falling with the electron... True (I think) because of equivalence principle, but I am not wholly sure. There is potentially the detectable difference because of the tidal nature of the divergent field. What also seems to be true is that a non-charged body falls at the same rate. The challenge, debated in the Ericsen paper, is to explain how the charge can radiate energy, with no apparent limit, yet not feel the reaction force from the radiation i.e. why does it not fall more slowly?

An electron orbiting a star will emit radiation to a far away observer stationary with respect to the star but not to an observer at the center of the star... Not true. An electron orbiting a star will not emit any radiation (to any observer), despite being accelerated by the gravitational centripetal force. It cannot or its orbit would decay and this does not occur. This is explained by the Eriksen paper but is very complex. The reason for the difference is to do with the boundary conditions which, in this case, are periodic (the electron starts and finishes its orbit on the same point).

This is reasoned from the derived equations but does not seem obvious to me.
 

Offline Ron Hughes

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« Reply #63 on: 18/02/2010 18:44:06 »
I think the problem is this, energy is only energy from an observers frame of reference. Suppose your out in space in your rocket ship. You see an electron parked outside your window. The electron starts accelerating and you see that it is emitting radiation. You turn on your engines and start to accelerate at the same rate. You notice that the radiation has magically disappeared.
 

Offline yor_on

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« Reply #64 on: 19/02/2010 08:07:20 »
"But I do not understand the action of an accelerating charge in a gravity field. And, it seems surprisingly, neither does anyone else"

I soo agree Graham :)
And I really like this discussion.

Like, how to define a 'free electron'.
One definition I saw was this..

"A free electron mean an isolated electron. If the electron is accelerating, it cannot be free or isolated, and there must be external fields acting on it."

But that is wrong in this case, isn't it? Gravity may look like a 'field' in that it is everywhere and decide all motions, from light to matter, but it's more like the stress lines in a jelly (aka SpaceTimes Geodesics) than a 'force' to me. Now, considering if 'gravity waves' is 'energy'?

I liked Vernons definition there "General Relativity does not offer such a close relationship, though; the gravitational field is simply a fixed region of curved space, and a gravitational wave is a moving ripple of space." But there is something strange with the concept. If it 'moves' relative me and, nota bene, against SpaceTimes geodesics shouldn't that take 'energy'?

So, it seems to fall back to if gravity can be seen as energy again :)
Or is there some other way to see it?
==

And how about inertia? Existing everywhere, reacting instantaneously on course change. doesn't that state something different than those 'gravity waves'. To me inertia is a 'refusal' to change, aka the 'jello' being SpaceTime 'braking' a proceeding/interaction by it's predefined geometry?
« Last Edit: 19/02/2010 08:21:29 by yor_on »
 

Offline graham.d

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« Reply #65 on: 19/02/2010 09:56:45 »
Ron, I think you should go through Eriksen's paper. If it were as simple as you state, I would not have any problem with the subject. I did not appreciate the complexity until digging into it. In your example, and to remove any magic, put the "parked" electron on another rocket ship so we can see what is providing the force to accelerate it. When the rocket ship accelerates with the electron, you will see the electron radiate and the result will be as the Larmor equation predicts. However, will the rocket ship's acceleration be very slightly reduced by the presence of the electron (compared with the presence of another particle of similar rest mass)? The (surprising) answer is "no". The energy for radiation seems to come from the mysterious (at least to me) Schott energy which seems to have no physical limit and can go infinitely negative in value!! This is what is being discussed in the Eriksen paper. The equations also have to be modified because it seems the rate of change of proper acceleration has a significant effect too.

If you accelerate your rocket to match the electron's acceleration, both you and the electron should be experiencing the same gravitational field as you are both existing in the same linearly accelerating frame. As electrons do not radiate in a constant gravitational field, then I agree, you will not observe radiation. I think you may see radiation at the start of the electron's and your acceleration because of the term associated with the derivative of the acceleration, even if you are in the same frame. The maths is too hard, but I think this would be the case.

 

Offline yor_on

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« Reply #66 on: 19/02/2010 20:54:51 »
Here is one definition of Schott energy

And this Generalization of the Schott energy in electrodynamic radiation theory

--Quote--

It arises from the fact that the power supplied by an external force to a charged particle not only contributes to the energy radiated (acceleration fields) but also to the velocity fields. This feature is not connected with the well-known eficiencies of the Abraham-Lorentz theory (runaway solutions, etc.). Previous discussions of the Schott energy arose in the context of the Abraham-Lorentz equation of motion for a radiating electron. In this paper we define a (generalized) Schott energy that is applicable not only to the Abraham-Lorentz theory but to all theories of a radiating electron.

--End of quote--

But I agree, it seems weird, and it can't relate to gravity wells?
If we haven't suddenly defined gravity as some sort of electromagnetic energy while I was sleeping?
« Last Edit: 19/02/2010 21:13:10 by yor_on »
 

Offline graham.d

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« Reply #67 on: 19/02/2010 21:46:19 »
It is good you found some more papers on the subject, Yor_on, as all the ones I found were ones you have to pay for. Looking through these ones briefly it looks like the maths is much more manageable. I'm away tomorrow but will see if I can get some time between chores on Sunday. It's a hard life :-)
 

Offline yor_on

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« Reply #68 on: 19/02/2010 22:33:08 »
Well Graham candles and nighttime is still free (okay, not the candles perhaps), just inform your spouse that you have more important things to do. And remember to duck when she delivers her 'answer' :)
 

Offline yor_on

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« Reply #69 on: 16/04/2010 12:38:13 »
Ah Farsight?
"There's some interesting comments re Einstein and gravity, about point particles, and even the near field aka evanescent wave:"

Evanescent waves again?

Anyway, I have a question. At another place, not far away in cyberspace, I wrote this.

"As for dropping an electron into a gravitational field?

That one gives me a headache :) Consider it empty on other particles, just you and that electron 'free falling' down some ??? No 'static' EM fields around, just that electron. Would it then radiate? And then have a far observer at rest versus the gravitational field 'accelerating' you. Would they see the same?

Why?"

As I'm still not sure how to see it. I got the answer that this question already was solved, pointing me to the Liénard–Wiechert potential by Alfred-Marie Liénard in 1898 and independently by Emil Wiechert in 1900 To my eyes it is not solved, but I'm curious to what degree any one else will agree or disagree with me? So take a look at the link and see if it defines the answer to my question here, as well as to the other ones discussed in this thread?
 

Offline graham.d

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« Reply #70 on: 18/04/2010 18:22:32 »
I don't think this solves the issue with an accelerating charge as it is only consistent with Maxwell's equations (and special relativity) not necessarily GR. This whole issue is tied in with Action at a Distance problems and work started, but not finished, by Feynman with contributions by Wheeler (and some earlier work by Fokker) I think. Surprisingly perhaps, I don't think there is any general consensus view on the behaviour of accelerating charges in all conditions - at least I have not found one in my limited research. The concept (with Action at a distance theories) requires retarded and advanced waves and interaction of a charge with the rest of the universe so it is somewhat complex and depends crucially on the model of the universe too. Even Feynman found it a daunting task!

The issue with who sees what with a gravitationally accelerating charge should be simple; it is certainly an easy problem to state, but a wholly GR consistent answer is not easy to find.
 

Offline Pmb

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?Accelerating a charged particle
« Reply #71 on: 19/04/2010 12:26:44 »
Quote from: Ron Hughes
An accelerated charged particle emits radiation with respect to a stationary observer but if the observer is accelerated along with the particle does the observer still detect the particle emitting radiation?.
That depends on who is doing the observing. Radiation emitted by a charged particle can only be detected by an instruement when the instrument itself is not co-moving with respect to the rest frame of the charged article.

This point was addressed in The American Journal of physics in the

“Radiation from an Accelerated Charge and the Principle of equivalence,” A. Kovetz and G.E. Tauber, 37(4), April 1969

The abstract reads
Quote
The connection between an accelerated charge and one at rest in a (weak) gravitational field is discussed in accordance with the principle of equivalence . For that purpose, the fields produced by a freely falling charge and a supported one (i.e., at rest in a gravitational field) are transformed to the rest-frame of the observer, who may be similarly supported or freely falling. A nonvanishing energy flux is found only if the charge is freely and the observer supported. or vice versa. This agrees with previously established results.
 

Offline graham.d

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« Reply #72 on: 19/04/2010 15:34:14 »
This may be right, Pmb, but it is still unclear to me in every circumstance. A distant observer would also be a free falling observer (though not falling very fast) and he would not see the charged particle radiate any more than someone falling next to the particle. Fair enough. Orbiting is also free falling so would the orbit decay? An observer on the ground should see that the particle is in free fall, and he is not, so he should see it emit shouldn't he? If it is emitting, the orbit should decay. But wait a minute. Someone in orbit with the particle (or a long way off) will not see it emit, so will the particle decay despite there being no apparent reason for it to do so?
 

Offline yor_on

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« Reply #73 on: 19/04/2010 17:54:26 »
Pmb how is a electron thought to radiate?
Is it orthogonally (perpendicular) to its motion?
It seems to me that this should be the way?
But I may easily be wrong there.

(It's a EM radiation, right?)

If so, when are the instrument co-moving with respect to that rest frame of the charged article. Is it when having the same motion, being 'at rest' with the electrons motion, or is it when moving orthogonally, at rest with the radiation?
===

Or is this just plain silly :)
===

Those damn* eleclons, now they are taking my interest away from beloved photons too.
But that's all there is right?

Photons and those, whatever they were called?
I mean, how many do we need?
==

The electron is described as having at all times, moving or unmoving, a 'field', as follows by Maxwell's equations, called the 'coulomb field'. That field seems to 'stretch' in all directions around our electron. When the electron moves relative an observer, there will come two new fields, one is the electric current that is the 'electron' in itself while moving? And surrounding that a magnetic field. Together those two new fields are named the dynamic electric field. " It's useful to regard the dynamic electric field as the sum of two separate fields, one of which is in phase with the magnetic field and the other 90 degrees out of phase. We will call the in-phase component the radiation field and the out-of-phase component the induction field. It is the radiation field that carries energy from an antenna into the surrounding universe" 

So what exactly is it that radiates in our thought example before? And how does it radiate? Doesn't it need a magnetic field to be pre-existent before it can do so? If I assume a perfect vacuum and leave this electron free falling into a gravitational 'well' will this description still be correct? meaning that it will create a current, as well as a magnetic field??

Awh, I should have stayed with photons :)
« Last Edit: 19/04/2010 20:07:30 by yor_on »
 

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